Problem: You have found the following ages (in years) of 5 lizards. Those lizards were randomly selected from the 29 lizards at your local zoo: $ 1,\enspace 1,\enspace 2,\enspace 4,\enspace 1$ Based on your sample, what is the average age of the lizards? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 29 lizards, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{1 + 1 + 2 + 4 + 1}{{5}} = {1.8\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {0.64} + {0.64} + {0.04} + {4.84} + {0.64}} {{5 - 1}} $ {s^2} = \dfrac{{6.8}}{{4}} = {1.7\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{1.7\text{ years}^2}} = {1.3\text{ years}} $ We can estimate that the average lizard at the zoo is 1.8 years old. There is also a standard deviation of 1.3 years.